Bob Mackinnon

1435 Opposite 1NT

In a previous blog we employed the expected numbers of total trumps as a guide for competing with a weak 4441 hand opposite a strong NT opening bid. We noted that a second nonforcing bid by responder is required in order to stop at a low level if the initial reply to Stayman doesn’t uncover a major 4-4 fit. Here we look at the case of responder’s holding a 1=4=3=5 hand and less than game-forcing values.

Again we calculate the a posteriori probabilities under the assumption that the opener holds a hand that is 4333, 4432 or 5332 with a 5-card minor. From these we obtain the probabilities of the various divisions of sides, hence probabilities of the number of total trumps.

Total Tricks

1=4=4=4

1=4=3=5

15

9%

8%

16

36%

29%

17

31%

37%

18

20%

15%

19

4%

9%

20

0%

1%

The most pronounced difference between the 2 hand shapes is that for 1=4=3=5 the most frequent number of total trumps is 17, not 16, as the a priori probabilities predict. This is expected, as for 1=4=3=5 the difference between the longest suit and the shortest suit is 4, the maximum contribution to total trumps being 17 (13+4) instead of 16 (13+3). Indeed, there is a 2/3 chance of 16 or 17 total trumps and an 80% chance of having 16, 17 or 18 total trumps. This encourages activity even on weak hands that carry the auction to the 3-level.

The Replies to Stayman

The normal replies to Stayman are limited to 3 calls: 2, 2, and 2. There is little provision for finding partials in a minor suit. In 3NT it is usually the majors that represent the first line of defence and the minors that produce the bulk of the tricks. The minors come into their own in the part score deals, usually played at the 3-level.

With the 1=4=3=5 distribution the 3 standard replies occur equally 1/3 of the time. The 2 reply discloses the 4-4 major fit immediately and doesn’t deny opener also holding 4 spades. The 2 reply will deliver 3 hearts 4 out of 5 times, so a nonforcing 2/2 may land responder in a 4-3 fit. He may not be left to play there.

2NT Asking after a 2 Reply

The number of forcing bids one requires depends on whether one is describing one’s hand or asking partner to describe his. One forcing bid is all that is needed when one player has a good appreciation of what to expect in his partner’s hand, as here. The five possibilities for the opener are ranked according to frequency as follows.

Case

Shape

Sides

Occurrence

1

4=3=3=3

5=7=6=8

28%

2

4=3=4=2

5=7=7=7

24%

3

4=3=2=4

5=7=5=9

21%

4

4=2=4=3

5=6=7=8

15%

5

4=2=3=4

5=6=6=9

13%

As responder holds 5 clubs, he can bid 3 to play, indicating invitational values. Opener may pass or take a shot at 3NT if he has a maximum with club support. That being the case, responder needs a forcing bid for his better hands and 2NT can be used in that capacity. Opener is asked to bid 3 or 3, the latter case only when his diamonds are longer than his clubs. This reply scheme is necessary in case responder has 1=4=4=4 shape and wants to play in the better minor suit fit. Responder may take further action if he has a good hand, perhaps by bidding 3, forcing. The meanings of these sequences are different from the current standard practice, where, for example, 3/2 would be a slam try with long clubs and 4 hearts. (I don’t recall ever having reached slam by that route, but, who knows? it could happen tomorrow.)

Having reached this point one cannot expect the opponents to enter the auction and rescue one from a bad contract, so it is important that one have a suitable hand for declarer play in 3 of a minor. What might that look like? As 3 would be a descriptive bid, inviting further action by opener, it is best if the hand conforms to expectations, that is, no top honour in spades, honours in clubs. Hearts needn’t be strong, but better than diamonds, as hearts may sometimes be required to be trumps in a 4-3 fit. Here is a fitting example from a computer search.

W
West
KJ97
AJ
J754
A105
 
E
East
10
Q1032
932
KQ432
West
East
1NT
2
2
3
3NT
Pass

Game was made on defensive mistiming when AKQ were cashed early setting up the J. One often sees this sort of thing at the local club. The opening leader may have been eager to cash his tricks before they disappeared on the clubs. Wait a minute, can a computer feel that way?

Here is another computer deal where it paid to bid aggressively on a less suitable distribution of HCPs. Fortune favours the bold, and sometimes everything falls into place.

 
E-W
South
N
North
4
K1042
K85
J8732
 
W
West
A3
J853
Q7632
Q4
 
E
East
Q108762
Q9
AJ
965
 
S
South
KJ95
A76
1094
AK10
 
W
West
N
North
E
East
S
South
1NT
Pass
2
Pass
2
Pass
3
Pass
3NT

The North hand has a meagre club suit, so not ideal for an action that might lead to a 3 contract, but there was a chance of playing in hearts. The K should be the K. Given the hope of a makeable game by North’s 3 escape, South feels that 3NT is quite possible. Note he holds 6 controls, worth 20 equivalent points, with great club support.

West led the 3, ducked to the J. The A was cashed and the 5 followed. West won the 9 with the A and returned a passive diamond won by dummy’s king. Declarer played the AK, dropping the Q and made his 9th trick in spades. This may appear miraculous, but Deep Finesse shows us that 3NT is unbeatable on any lead, and in some sequences declarer can afford to lose the club finesse. So, it is not correct to state the contract depended on dropping the Q doubleton.

Some will be concerned that 3NT was an option on a mere 22 HCP, but one should be aware of the nature of those points: 2 aces, 4 kings, and no queens. Eight control points with a long suit to run are worthy of game consideration. One may think NS were very lucky here, but I think that a game that can make on any opening lead is worth bidding, and that it is up to the bidding system to get the user there. A system based solely on HCPs is not doing its job.

The 2 Reply

If the opening bidder doesn’t hold a 4-card major his response will be 2, so responder will know immediately that the opponents hold at least a 9-card spade fit. It is a situation where several live possibilities exist. The final contract will be resolved in an atmosphere of uncertainty with the responder in the best position to resolve it. Uncertainty is manifest in the multiplicity of the possible division of sides that remain.

Case

Shape

Sides

Occurrence

1

3=3=4=3

4=7=7=8

23%

2

3=3=3=4

4=7=6=9

17%

3

3=3=5=2

4=7=8=7

14%

4

3=2=5=3

4=6=8=8

12%

 

 

 

 

5

2=3=4=4

3=7=7=9

9%

6

2=3=5=3

3=7=8=8

8%

7

3=2=3=5

4=6=6=10

6%

8

3=4=2=5

4=7=5=10

5%

9

2=3=3=5

3=7=6=10

4%

10

3=2=4=4

4=6=7=9

3%

It is an exciting moment for the responder as he, and only he, knows the opponents have a workable 9-card fit in spades. Passing 2 is not a good idea. There is a 1/3 chance of an 8-card fit, but such passivity may only set the LHO in motion. Bidding 3 is likely to find club support in dummy, and the better the support, the more likely it is that partner will push to game. Not so good. A nonforcing bid of 2 is a way to keep the ball rolling with all options alive. The overall chances of hitting a 4-3 heart fit are over 80%, but the door is left open for the opponents to make a move.

Here is an example from the computer where the opener bid an encouraging 2NT on the basis of his good controls in the majors and good fillers in the minors. That is the formula for success.

W
West
KQ2
AQ7
J107
QJ106
 
E
East
4
K1085
Q64
K8752
West
East
1NT
2
2
2
2NT
3NT

Responder was a maximum for an invitational bid. Spades were led, but with the A onside and the AK split, there was no problem making 10 tricks on sleepy defence. More often the opening pair will find themselves playing in a dicey 3, where the opponents can make 8 or 9 tricks in spades. Here is a rare 4=6=7=9 sides (Case 10).

W
West
K106
A3
Q643
AJ72
 
E
East
9
10965
J87
KQ862
West
East
1NT
2
2
3
Pass
 

Rather than making an ambiguous move in hearts responder bids where he lives hoping to steal the contract. Partner has exceptional support for clubs, but recognizes that a single stopper in either major may not be enough to assure 9 tricks. The opponents can make 9 tricks in spades against a porous defence, so even 4 off 1 would be a good result. The number of total tricks is 18.

In the next blog we shall investigate how to bid slams within the structure of nonforcing rebids by responder.


5 Comments

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Bob MacKinnonMarch 1st, 2016 at 7:05 pm

Thanks. My objective is to encourage the readers to think outside the bidding box and make up their own mind about whatever.

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